Python Monty Hall Simulation 没有给出预期的答案

Question

我在 Monty Hall 模拟中得到答案 0.5。

来自教科书：我们假设通过滚动三面骰子将汽车放在门后，这使得所有三个选择的可能性相同。 Monty 知道汽车在哪里，并且总是打开一扇门，后面有一只山羊。最后，我们假设如果 Monty 可以选择门（即参赛者选择了后面有汽车的门），他选择每扇门的概率为 1/2。玛丽莲显然希望她的读者认为游戏是以这种方式进行的。

Marilyn 的答案是 0.66，我想模拟这个答案，但是我得到了 0.5，不知道我的代码有什么问题。

n = 1000000
count = 0
for i in range(n):
    doors = [1,2,3]  
    # the inital doors that monty can choose
    monty_choose = [1,2,3]
    # suppose the car is behind door 1
    car = 1   
    # monty cannot choose the door that has car
    monty_choose.remove(car)    
    ichoose = random.choice(doors)  
    if ichoose in monty_choose:
        # monty cannot choose the door i select
        monty_choose.remove(ichoose)  
        monty = random.choice(monty_choose)  
    else:
        monty = random.choice(monty_choose)
    # i cannot choose the door that monty chose  
    doors.remove(monty)   
    s = random.choice(doors)
    if s == car:
        count = count + 1 
print(count/n)

Answer 1

doors = [1,2,3]  # total doors
i_choose = [1,2,3] # the inital doors that I can choose
car = 1   # suppose the car is behind door 1
host_choose = [2,3] # the empty doors the host could show
n = 1000000
count = 0
car = 1
for i in range(n):
    # you can randomize car here, but remember to change host_choose accordingly
    i_choice = random.choice(doors) # I choose one door randomly
    if first_choice in host_choose:
        host_choose.remove(first_choice) # the host cannot open the chosen door
    host_choice = random.choice(host_choose) # the host shows that a door is empty
    i_choose.remove(host_choice)
    i_choose.remove(first_choice)
    # the goal is to show that always changing door results in 66% winrate
    i_choice = random.choice(i_choose) # this is actually a one-element list
    if i_choice == car:
        count = count + 1
    i_choose = [1,2,3]
    doors = [1,2,3]
    host_choose = [2,3]
print(count/n)

所以基本上，您混淆了选择的对象和时间：

A 在 [1, 2, 3] 选了一扇门
B 知道车在哪里，并露出一扇空门（A 没有选择）
A 现在可以选择保留或更换门

你的目标是证明换门有 0.66 的概率得到汽车。

Answer 2

您的代码可以在您到达最后一点之前找到。您正在随机选择门：


 s = random.choice(doors)
    if s == car:
        count = count + 1

当您想做的是换门时。您可以通过简单地删除您的第一个选择然后将列表索引为 0 来做到这一点。

    doors.remove(ichoose)
    
    if doors[0] == car:
        count = count + 1

完整代码和结果

import random

n = 1000000
count = 0
for i in range(n):
    doors = [1,2,3]  
    # the inital doors that monty can choose
    monty_choose = [1,2,3]
    # suppose the car is behind door 1
    car = 1   
    # monty cannot choose the door that has car
    
    monty_choose.remove(car)    
    ichoose = random.choice(doors)  
    if ichoose in monty_choose:
        # monty cannot choose the door i select
        monty_choose.remove(ichoose) 
        monty = random.choice(monty_choose)
    else:
        monty = random.choice(monty_choose)
    # i cannot choose the door that monty chose  
    doors.remove(monty)  
    
    
    doors.remove(ichoose)
    
    if doors[0] == car:
        count = count + 1 
        
        
print(count/n)
0.667145

Answer 3

您的代码计算 0.5 的概率仅仅是因为 s = random.choice(doors) 从汽车或山羊中选择的概率相同。

您的代码没有反映 Monty Hall 问题的工作原理。

如果参赛者做出一个选择并坚持那个选择，那么概率显然是0.33。你永远不允许 ichoose 坚持他们的选择。

不太明显的部分是参赛者可以改变他们的选择，那么概率是0.66。你永远不允许 ichoose 改变他们的选择。

Python Monty Hall Simulation 没有给出预期的答案

Python Monty Hall Simulation didn't give expected answer

python

logic

simulation